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Question: Find the equation of a circle passing through the points \((3, - 2),\,\,(2,0)\) and hence its centre...

Find the equation of a circle passing through the points (3,−2),  (2,0)(3, - 2),\,\,(2,0) and hence its centre lies on the line 2x−y=32x - y = 3?

Explanation

Solution

Here we have to find the equation of a circle which passes through the points (3,−2),  (2,0)(3, - 2),\,\,(2,0) and its centre lies on the line 2x−y=32x - y = 3. To find the equation of a circle we will use the standard or general equation of a circle which is given as (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}. In the above equation rr is the radius of a circle, (h,k)(h,k) are coordinates of its centre.

Complete step by step answer:
In the given question we have to find the equation of a circle which passes through the points (3,−2),  (2,0)(3, - 2),\,\,(2,0) and its centre lies on the line 2x−y=32x - y = 3. We know that the equation of a circle with centre (h,k)(h,k) is,
(x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
Since the circle passes through(3,−2)(3, - 2). We have
⇒(3−h)2+(−2−k)2=r2\Rightarrow {(3 - h)^2} + {( - 2 - k)^2} = {r^2}

Use the identity (a−b)2=a2+b2=2ab{(a - b)^2} = {a^2} + {b^2} = 2ab to solve the above equation.
⇒9+h2−6h+4+k2+4k=r2\Rightarrow 9 + {h^2} - 6h + 4 + {k^2} + 4k = {r^2}
⇒h2+k2−6h+4k+13=r2……(1)\Rightarrow {h^2} + {k^2} - 6h + 4k + 13 = {r^2} \ldots \ldots (1)
Since the circle passes through (−2,0)( - 2,0). We have
(−2−h)2+(0−k)2=r2{( - 2 - h)^2} + {(0 - k)^2} = {r^2}
Use the identity (a−b)2=a2+b2=2ab{(a - b)^2} = {a^2} + {b^2} = 2ab to solve the above equation.
⇒4+h2+4h+k2=r2\Rightarrow 4 + {h^2} + 4h + {k^2} = {r^2}
⇒h2+k2+4h+4=r2……(2)\Rightarrow {h^2} + {k^2} + 4h + 4 = {r^2} \ldots \ldots (2)

Subtracting equation (1)(1) from equation (2)(2). We get,
⇒h2+k2+4h+4−h2−k2+6h−4k−13=r2−r2\Rightarrow {h^2} + {k^2} + 4h + 4 - {h^2} - {k^2} + 6h - 4k - 13 = {r^2} - {r^2}
Simplifying the above equation. We get,
⇒10h−4k=9……(3)\Rightarrow 10h - 4k = 9 \ldots \ldots (3)
Since centre (h,k)(h,k) lies on the line 2x−y=32x - y = 3
So, we can write it as 2h−k=3……(4)2h - k = 3 \ldots \ldots (4)
Now we will solve equation (3)(3) and (4)(4) by elimination to find the value of (h,k)(h,k)
We have
10h−4k=9 ⇒2h−k=3 10h - 4k = 9 \\\ \Rightarrow 2h - k = 3 \\\

Multiplying the equation 2h−k=32h - k = 3 by (−4)( - 4). We have,
10h−4k=9 ⇒−8h+4k=−12 10h - 4k = 9 \\\ \Rightarrow - 8h + 4k = - 12 \\\
On adding both the equations. We get,
⇒2h=−3\Rightarrow 2h = - 3
⇒h=−32\Rightarrow h = - \dfrac{3}{2}
Putting the value of h=−32h = - \dfrac{3}{2} in equation (4)(4). We get,
⇒2×(−32)−k=3\Rightarrow 2 \times \left( {\dfrac{{ - 3}}{2}} \right) - k = 3
⇒−3−k=3\Rightarrow - 3 - k = 3
⇒k=−6\Rightarrow k = - 6

Now, let us find the radius of a circle by putting the value h=−32h = - \dfrac{3}{2} and k=−6k = - 6 in equation (1)(1). We get,
⇒(−32)2+(−6)2−6(−32)+4(−6)+13=r2\Rightarrow {\left( {\dfrac{{ - 3}}{2}} \right)^2} + {( - 6)^2} - 6\left( {\dfrac{{ - 3}}{2}} \right) + 4( - 6) + 13 = {r^2}
⇒(94)+36+9−24+13=r2\Rightarrow \left( {\dfrac{9}{4}} \right) + 36 + 9 - 24 + 13 = {r^2}
⇒(94)+34=r2\Rightarrow \left( {\dfrac{9}{4}} \right) + 34 = {r^2}
⇒9+1364=r2\Rightarrow \dfrac{{9 + 136}}{4} = {r^2}
⇒r2=1454\Rightarrow {r^2} = \dfrac{{145}}{4}
Put the value of (h,k)(h,k) and r2{r^2} in the standard equation of the circle. We get,
⇒(x+32)2+(y+6)2=1454\Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {(y + 6)^2} = \dfrac{{145}}{4}
∴(2x+3)2+4(y+6)2=145\therefore {(2x + 3)^2} + 4{(y + 6)^2} = 145

Hence, the equation of a circle which passes through the points (3,−2),  (2,0)(3, - 2),\,\,(2,0) and whose centre lie on the line 2x−y=32x - y = 3is (2x+3)2+4(y+6)2=145{(2x + 3)^2} + 4{(y + 6)^2} = 145.

Note: We can solve these types of problems by applying the formula of distance between two points. If (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) are the two points, then the distance between these points is given by the formula d=(x1−x2)2+(y1−y2)2d = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} .